Question

in a poll of 520 human resource professionals, 45.8% said that body piercings and tattoos were big personal grooming red flags. complete parts (a) through (d) below.
a. among the 520 human resource professionals who were surveyed, how many of them said that body piercings and tattoos were big personal grooming red flags?
238 (round to the nearest integer as needed)
b. construct a 99% confidence interval estimate of the proportion of all human resource professionals believing that body piercings and tattoos are big personal grooming red flags.
□ < p < □
(round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}$ is given as $0.458$, and the sample size $n = 520$.

Step2: Find critical value

For a $99\%$ confidence interval, the significance level $\alpha=1 - 0.99 = 0.01$, so $\alpha/2=0.005$. The critical value $z_{\alpha/2}$ is $z_{0.005} = 2.576$ (from standard normal distribution table).

Step3: Calculate margin of error

The margin of error $E$ is given by the formula $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substitute the values: $\hat{p}=0.458$, $n = 520$, $z_{\alpha/2}=2.576$.
First, calculate $\hat{p}(1 - \hat{p})=0.458\times(1 - 0.458)=0.458\times0.542 = 0.248236$.
Then, $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.248236}{520}}\approx\sqrt{0.000477377}\approx0.02185$.
$E = 2.576\times0.02185\approx0.0563$.

Step4: Calculate confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-ESubstitute $\hat{p}=0.458$ and $E = 0.0563$.
$0.458-0.0563 < p<0.458 + 0.0563$.

Answer:

$0.402 < p<0.514$